Finding the Exact Value of sin[2arcsin(3/5)] with Inverse Trig Functions

[page13]

Homework Statement

Find the exact value of the expression:

Homework Equations

sin[2arcsin(3/5)]

The Attempt at a Solution

I know you're supposed to use sin2x=2sinxcosx somehow but not sure how to start.

 

[Mark44]

Homework Statement

Find the exact value of the expression:

Homework Equations

sin[2arcsin(3/5)]

The Attempt at a Solution

I know you're supposed to use sin2x=2sinxcosx somehow but not sure how to start.

Start by letting u = arcsin(3/5). Then your expression is sin(2u) = ?

It will be helpful to draw a right triangle where u is one of the acute angles. Label the sides and hypotenuse so that sin(u) = 3/5.

 

[page13]

Ah, OK! 24/25 correct?